NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4

Exercise 2.4 of Class 9 Maths Chapter 2 - Polynomials introduces the essential concept of the Remainder Theorem. In this exercise, you will learn to use the Remainder Theorem to find the remainder when a polynomial is divided by a linear polynomial. This concept is helpfull when solving larger algebra problems later on.

The Solutions in this exercise are from the NCERT textbooks and practicing them gives you a good chance of scoring better marks in your exams and gives you a clear understanding of the key concepts of polynomials.

To help your learning, we have provided step by step NCERT Solutions for Exercise 2.4 in this article. You can easily download a free PDF of our Practice Solutions to use to practice anytime to develop confidence and be better prepared for exams!

1.0Download NCERT Solutions of Class 9 Maths Chapter 2 Polynomials Exercise 2.4: Free PDF

Exercise 2.4 helps you understand the applications of Remainder Theorem for easy calculation. The NCERT Solutions of Class 9 Maths Chapter 2, provides a step by step solution to the questions of this exercise. Download the free PDF of the solutions from below:

2.0NCERT Exercise Solutions Class 9 Chapter 2- Polynomials: All Exercises 

3.0NCERT Class 9 Maths Chapter 2 Exercise 2.4 : Detailed Solutions

1. Use suitable identities to find the following products :

(i) (x+4)(x+10)

(ii) (x+8)(x−10)

(iii) (3x+4)(3x−5)

(iv) (y²+3/2)(y²−3/2)

(v) (3−2x)(3+2x)

Sol. (i) (x+4)(x+10) [using identity (x+a)(x+b) = x²+(a+b)x+ab]

= x²+(4+10)x+(4)(10)

= x²+14x+40

(ii) (x+8)(x−10) [using identity (x+a)(x+b) = x²+(a+b)x+ab]

= (x+8)(x+(−10))

= x²+{8+(−10)}x+8(−10)

= x²−2x−80

(iii) (3x+4)(3x−5) [using identity (x+a)(x+b) = x²+(a+b)x+ab, where x is 3x]

= (3x+4)(3x+(−5))

= (3x)²+{4+(−5)}(3x)+4(−5)

= 9x²−3x−20

(iv) (y²+3/2)(y²−3/2) [using identity (a+b)(a−b)=a²−b²]

Let a=y² and b=3/2.

= (y²)²−(3/2)²

= y⁴−9/4

(v) (3−2x)(3+2x) [using identity (a−b)(a+b)=a²−b²]

= (3)²−(2x)²

= 9−4x²

2. Evaluate the following products without multiplying directly:

(i) 103×107

(ii) 95×96

(iii) 104×96

Sol. (i) 103×107

= (100+3)×(100+7) [using identity (x+a)(x+b) = x²+(a+b)x+ab]

= (100)²+(3+7)(100)+(3)(7)

= 10000+1000+21

= 11021

Alternate solution: 103×107 = (105−2)×(105+2) = (105)²−(2)² = 11025−4 = 11021.

(ii) 95×96

= (90+5)×(90+6) [using identity (x+a)(x+b) = x²+(a+b)x+ab]

= (90)²+(5+6)90+(5)(6)

= 8100+11(90)+30

= 8100+990+30

= 9120

(iii) 104×96

= (100+4)×(100−4) [using identity (a+b)(a−b)=a²−b²]

= (100)²−(4)²

= 10000−16

= 9984

3. Factorise the following using appropriate identities:

(i) 9x²+6xy+y²

(ii) 4y²−4y+1

(iii) x²−y²/100

Sol. (i) 9x²+6xy+y² [using identity (a+b)²=a²+2ab+b²]

= (3x)²+2(3x)(y)+(y)²

= (3x+y)²

= (3x+y)(3x+y)

(ii) 4y²−4y+1 [using identity (a−b)²=a²−2ab+b²]

= (2y)²−2(2y)(1)+(1)²

= (2y−1)²

= (2y−1)(2y−1)

(iii) x²−y²/100 [using identity (a²−b²)=(a+b)(a−b)]

= x²−(y/10)²

= (x+y/10)(x−y/10)

4. Expand each of the following using suitable identities :

(i) (x+2y+4z)²

(ii) (2x−y+z)²

(iii) (−2x+3y+2z)²

(iv) (3a−7b−c)²

(v) (−2x+5y−3z)²

(vi) (1/4 a−1/2 b+1)²

Sol. (Using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca)

(i) (x+2y+4z)²

= (x)²+(2y)²+(4z)²+2(x)(2y)+2(2y)(4z)+2(4z)(x)

= x²+4y²+16z²+4xy+16yz+8zx

(ii) (2x−y+z)²

= (2x)²+(−y)²+(z)²+2(2x)(−y)+2(−y)(z)+2(z)(2x)

= 4x²+y²+z²−4xy−2yz+4zx

(iii) (−2x+3y+2z)²

= (−2x)²+(3y)²+(2z)²+2(−2x)(3y)+2(3y)(2z)+2(2z)(−2x) (Note: corrected 2nd last term as per identity)

= 4x²+9y²+4z²−12xy+12yz−8xz

(iv) (3a−7b−c)²

= (3a)²+(−7b)²+(−c)²+2(3a)(−7b)+2(−7b)(−c)+2(−c)(3a) (Note: corrected 2nd last term as per identity)

= 9a²+49b²+c²−42ab+14bc−6ac

(v) (−2x+5y−3z)²

= (−2x)²+(5y)²+(−3z)²+2(−2x)(5y)+2(5y)(−3z)+2(−3z)(−2x)

= 4x²+25y²+9z²−20xy−30yz+12xz

(vi) (1/4 a−1/2 b+1)²

= (1/4 a)²+(−1/2 b)²+(1)²+2(1/4 a)(−1/2 b)+2(−1/2 b)(1)+2(1)(1/4 a)

= (1/16)a²+(1/4)b²+1−(1/4)ab−b+(1/2)a

5. Factorise :

(i) 4x²+9y²+16z²+12xy−24yz−16xz

(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz

Sol. (i) 4x²+9y²+16z²+12xy−24yz−16xz [using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca]

Notice the negative terms involving z. This means z must be negative in the factored form.

= (2x)²+(3y)²+(−4z)²+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)

= (2x+3y−4z)²

= (2x+3y−4z)(2x+3y−4z)

(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz [using identity (a+b+c)²=a²+b²+c²+2ab+2bc+2ca]

Notice the negative terms involving x and xz. This means x must be negative in the factored form.

= (−√2x)²+(y)²+(2√2z)²+2(−√2x)(y)+2(y)(2√2z)+2(2√2z)(−√2x)

= (−√2x+y+2√2z)²

6. Write the following cubes in expanded form :

(i) (2x+1)³

(ii) (2a−3b)³

(iii) (3/2 x+1)³

(iv) (x−2/3 y)³

Sol. (Using identity (a+b)³=a³+b³+3ab(a+b) and (a−b)³=a³−b³−3ab(a−b))

(i) (2x+1)³

= (2x)³+(1)³+3(2x)(1)(2x+1)

= 8x³+1+6x(2x+1)

= 8x³+1+12x²+6x

= 8x³+12x²+6x+1

(ii) (2a−3b)³

= (2a)³−(3b)³−3(2a)(3b)(2a−3b)

= 8a³−27b³−18ab(2a−3b)

= 8a³−27b³−36a²b+54ab²

(iii) (3/2 x+1)³

= (3/2 x)³+(1)³+3(3/2 x)(1)(3/2 x+1)

= (27/8)x³+1+(9/2)x(3/2 x+1)

= (27/8)x³+1+(27/4)x²+(9/2)x

= (27/8)x³+(27/4)x²+(9/2)x+1

(iv) (x−2/3 y)³

= x³−(2/3 y)³−3x(2/3 y)(x−2/3 y)

= x³−(8/27)y³−2xy(x−2/3 y)

= x³−(8/27)y³−2x²y+(4/3)xy²

7. Evaluate the following using suitable identities :

(i) (99)³

(ii) (102)³

(iii) (998)³

Sol. (i) (99)³

= (100−1)³ [using (a−b)³=a³−b³−3ab(a−b)]

= (100)³−(1)³−3(100)(1)(100−1)

= 1000000−1−300(99)

= 1000000−1−29700

= 970299

(ii) (102)³

= (100+2)³ [using (a+b)³=a³+b³+3ab(a+b)]

= (100)³+(2)³+3(100)(2)(100+2)

= 1000000+8+600(102)

= 1000000+8+61200

= 1061208

(iii) (998)³

= (1000−2)³ [using (a−b)³=a³−b³−3ab(a−b)]

= (1000)³−(2)³−3(1000)(2)(1000−2)

= 1000000000−8−6000(998)

= 1000000000−8−5988000

= 994011992

8. Factorise each of the following :

(i) 8a³+b³+12a²b+6ab²

(ii) 8a³−b³−12a²b+6ab²

(iii) 27−125a³−135a+225a²

(iv) 64a³−27b³−144a²b+108ab² (Note: the original text had 108a² which seems like a typo, assuming 108ab² for consistency with (a-b)³)

(v) 27p³−1/216−(9/2)p²+(1/4)p

Sol. (i) 8a³+b³+12a²b+6ab² [using identity (a+b)³=a³+b³+3a²b+3ab² = a³+b³+3ab(a+b)]

= (2a)³+(b)³+3(2a)²(b)+3(2a)(b)²

= (2a)³+(b)³+3(2a)(b)(2a+b)

= (2a+b)³

= (2a+b)(2a+b)(2a+b)

(ii) 8a³−b³−12a²b+6ab² [using identity (a−b)³=a³−b³−3a²b+3ab²]

= (2a)³+(−b)³+3(2a)²(−b)+3(2a)(−b)²

= (2a−b)³

(iii) 27−125a³−135a+225a² [Rearrange to a³−b³−3a²b+3ab² form: (3)³−(5a)³+225a²−135a ]

= (3)³−(5a)³+3(3)(5a)(−5a+3) (incorrect grouping in solution)

Correct interpretation: 27−125a³−135a+225a² = 3³−(5a)³−3(3)²(5a)+3(3)(5a)²

= 3³−(5a)³−3(9)(5a)+3(3)(25a²)

= 3³−(5a)³−135a+225a²

This matches the form of (a−b)³ = a³−b³−3a²b+3ab².

So, it is (3−5a)³.

(iv) 64a³−27b³−144a²b+108ab² [using identity (a−b)³=a³−b³−3a²b+3ab²]

= (4a)³−(3b)³−3(4a)²(3b)+3(4a)(3b)²

= (4a)³−(3b)³−3(16a²)(3b)+3(4a)(9b²)

= (4a)³−(3b)³−144a²b+108ab²

So, it is (4a−3b)³.

(v) 27p³−1/216−(9/2)p²+(1/4)p [using identity (a−b)³=a³−b³−3a²b+3ab²]

= (3p)³−(1/6)³−3(3p)²(1/6)+3(3p)(1/6)²

= (3p)³−(1/6)³−3(9p²)(1/6)+3(3p)(1/36)

= (3p)³−(1/6)³−(27/6)p²+(9/36)p

= (3p)³−(1/6)³−(9/2)p²+(1/4)p

So, it is (3p−1/6)³.

= (3p−1/6)(3p−1/6)(3p−1/6)

Verify:

(i) x³+y³=(x+y)(x²−xy+y²)

(ii) x³−y³=(x−y)(x²+xy+y²)

Sol. (i) We know (x+y)³=x³+y³+3xy(x+y)

Rearranging for x³+y³:

x³+y³=(x+y)³−3xy(x+y)

Factor out (x+y):

x³+y³=(x+y)[(x+y)²−3xy]

Expand (x+y)²:

x³+y³=(x+y)(x²+2xy+y²−3xy)

Simplify:

x³+y³=(x+y)(x²−xy+y²) (Verified)

(ii) We know (x−y)³=x³−y³−3xy(x−y)

Rearranging for x³−y³:

x³−y³=(x−y)³+3xy(x−y)

Factor out (x−y):

x³−y³=(x−y)[(x−y)²+3xy]

Expand (x−y)²:

x³−y³=(x−y)(x²+y²−2xy+3xy)

Simplify:

x³−y³=(x−y)(x²+y²+xy) (Verified)

9. Factorise each of the following :

(i) 27y³+125z³

(ii) 64m³−343n³

Sol. (i) 27y³+125z³ [using identity a³+b³=(a+b)(a²−ab+b²)]

= (3y)³+(5z)³

= (3y+5z){(3y)²−(3y)(5z)+(5z)²}

= (3y+5z)(9y²−15yz+25z²)

(ii) 64m³−343n³ [using identity a³−b³=(a−b)(a²+ab+b²)]

= (4m)³−(7n)³

= (4m−7n){(4m)²+(4m)(7n)+(7n)²}

= (4m−7n)(16m²+28mn+49n²)

Factorise: 27x³+y³+z³−9xyz

Sol. 27x³+y³+z³−9xyz [using identity a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)]

Here, a=3x, b=y, c=z.

= (3x)³+(y)³+(z)³−3(3x)(y)(z)

= (3x+y+z)((3x)²+(y)²+(z)²−(3x)(y)−(y)(z)−(z)(3x))

= (3x+y+z)(9x²+y²+z²−3xy−yz−3zx)

Verify that x³+y³+z³−3xyz = (1/2)(x+y+z) [(x−y)²+(y−z)²+(z−x)²]

Sol. Start from the R.H.S.:

(1/2)(x+y+z)[(x−y)²+(y−z)²+(z−x)²]

Expand the squares:

= (1/2)(x+y+z)[(x²−2xy+y²)+(y²−2yz+z²)+(z²−2zx+x²)]

Combine like terms inside the square brackets:

= (1/2)(x+y+z)[2x²+2y²+2z²−2xy−2yz−2zx]

Factor out 2 from the square brackets:

= (1/2)(x+y+z)2(x²+y²+z²−xy−yz−zx)

Cancel the 1/2 and 2:

= (x+y+z)(x²+y²+z²−xy−yz−zx)

This is the standard identity for x³+y³+z³−3xyz.

= x³+y³+z³−3xyz (Verified)

If x+y+z=0, show that x³+y³+z³=3xyz.

Sol. We know the identity:

x³+y³+z³−3xyz=(x+y+z)(x²+y²+z²−xy−yz−zx)

Given: x+y+z=0. Substitute this into the identity:

x³+y³+z³−3xyz = (0)(x²+y²+z²−xy−yz−zx)

x³+y³+z³−3xyz = 0

Therefore, x³+y³+z³=3xyz.

10. Without actually calculating the cubes, find the value of each of the following :

(i) (−12)³+(7)³+(5)³

(ii) (28)³+(−15)³+(−13)³

Sol. (Using the identity: if a+b+c=0, then a³+b³+c³=3abc)

(i) Let a=−12, b=7, c=5.

Check if a+b+c=0: −12+7+5 = −12+12 = 0.

Since a+b+c=0, then a³+b³+c³=3abc.

(−12)³+(7)³+(5)³ = 3(−12)(7)(5)

= 3(−420) = −1260.

(ii) Let a=28, b=−15, c=−13.

Check if a+b+c=0: 28+(−15)+(−13) = 28−15−13 = 28−28 = 0.

Since a+b+c=0, then a³+b³+c³=3abc.

(28)³+(−15)³+(−13)³ = 3(28)(−15)(−13)

= 3(28)(195)

= 84 × 195 = 16380.

11. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :

(i) Area = 25a²−35a+12

(ii) Area = 35y²+13y−12

Sol. (Area of rectangle = Length × Breadth. Factorise the quadratic expression)

(i) Area = 25a²−35a+12

To factorise, find two numbers whose product is 25×12=300 and sum is -35. These are -20 and -15.

= 25a²−20a−15a+12

= 5a(5a−4)−3(5a−4)

= (5a−3)(5a−4)

Here, Length = 5a−3, Breadth = 5a−4 (or vice versa).

(ii) Area = 35y²+13y−12

To factorise, find two numbers whose product is 35×(−12)=−420 and sum is 13. These are 28 and -15.

= 35y²+28y−15y−12

= 7y(5y+4)−3(5y+4)

= (5y+4)(7y−3)

Here, Length = 5y+4, Breadth = 7y−3 (or vice versa).

12. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x²−12x

(ii) Volume: 12ky²+8ky−20k

Sol. (Volume of cuboid = Length × Breadth × Height. Factorise the expression into three factors)

(i) Volume = 3x²−12x

Factor out the common term 3x:

= 3x(x−4)

So, the dimensions are 3 units, x units and (x−4) units.

(ii) Volume = 12ky²+8ky−20k

Factor out the common term 4k:

= 4k(3y²+2y−5)

Now factor the quadratic expression (3y²+2y−5). Find two numbers whose product is 3×(−5)=−15 and sum is 2. These are 5 and -3.

= 4k(3y²+5y−3y−5)

= 4k{y(3y+5)−1(3y+5)}

= 4k(3y+5)(y−1)

So, the dimensions of the cuboid are 4k, (3y+5), and (y−1).

4.0Key Features and Benefits Class 9 Maths Chapter 2 Number System: Exercise 2.4

• Exercise 2.4 has questions that are simple and clear related to the concept of the Remainder Theorem.
• All problems are from the NCERT text book and thus makes scoring easier.
• These topics build a strong foundation for advanced algebraic division, which is useful in higher classes and some competitive exams like the olympiads.
• Regularly practicing these problems helps provide the needed confidence for the final exams.
• This also helps with sharpening your thinking ability and problem solving skills.